Answer:
The answer is (C)
Let’s run the algorithm on a small input to see the working process.
Let say we have an array {3, 4, 1, 5, 2, 7, 6}. MAX = 7
- Now for i=0, i < 7/2, here we exchange the value at ith index with value at (MAX-i-1)th index.
- So the array becomes {6, 4, 1, 5, 2, 7, 3}. //value at 0th index =3 and value at (7-0-1)th index is 6.
- Then for i=1, i < 7/2, the value at index 1 and (7-1-1)=5 are swapped.
- So the array becomes {6, 7, 1, 5, 2, 4, 3}.
- Then for i=2, i < 7/2, the value at index 2 and (7-2-1)=4 are swapped.
- So the array becomes {6, 7, 2, 5, 1, 4, 3}.
- Then for i=3, i not < 7/2, so stop here.
- Now the current array is {6, 7, 2, 5, 1, 4, 3} and the previous array was{3, 4, 1, 5, 2, 7, 6}.
Explanation:
So from the above execution, we got that the program reverses the numbers stored in the array.
Add the following constants in the class definition, before the main method:
private final static float START_COST = 1.5f;
private final static float HIGH_TARIFF = 0.5f;
private final static float LOW_TARIFF = 0.25f;
private final static int FIXED_MINS = 2;
private final static int HIGH_TARIFF_MINS = 10;
Then add this to the main method you already had:
float price = START_COST;
if (x > FIXED_MINS) {
if (x <= HIGH_TARIFF_MINS) {
price += HIGH_TARIFF*(x-FIXED_MINS);
}
else {
price += HIGH_TARIFF*(HIGH_TARIFF_MINS-FIXED_MINS)
+ LOW_TARIFF*(x-HIGH_TARIFF_MINS);
}
}
System.out.printf("A %d minute call costs %.2f", x, price);
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I will hope this helps :p
